The magnetic moment of [MnX_4]^{2-} is 5.9 , | vedclass

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Question

(A) The magnetic moment is given as $5.9 \, BM$. Using the spin-only formula $\mu = \sqrt{n(n+2)} \, BM$,where $n$ is the number of unpaired electrons

Vedclass · Accepted Answer

tetrahedral

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(A) The magnetic moment is given as $5.9 \, BM$. Using the spin-only formula $\mu = \sqrt{n(n+2)} \, BM$,where $n$ is the number of unpaired electrons,we find $n = 5$. In $[MnX_4]^{2-}$,the oxidation state of $Mn$ is $+2$. The electronic configuration of $Mn^{2+}$ is $[Ar] 3d^5$. Since there are $5$ unpaired electrons,the $d$-orbitals are $d^5$ (high spin). For a $d^5$ configuration with $4$ ligands,the geometry is tetrahedral because the halide ligands are weak field ligands,leading to $sp^3$ hybridization.

The magnetic moment of $[MnX_4]^{2-}$ is $5.9 \, BM$. The geometry of the complex ion is: ($X =$ monodentate halide ion)

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